chapter 6 – work and energyuregina.ca/~barbi/academic/phys109/2010/notes/lecture-16.pdf ·...

Chapter 6 – Work and Energy

Chapter 6

• Work Done by a Constant Force

• Kinetic Energy, and the Work-Energy Principle

• Potential Energy• Potential Energy

• Conservative and Nonconservative Forces

• Mechanical Energy and Its Conservation.

Recalling Recalling LastLast LectureLectureRecalling Recalling LastLast LectureLecture

Potential energy is the ability of an object todo work.

Gravitational potential energy is

Potential EnergyPotential Energy

(6-9)

(6-10)

The force needed to compress or stretch a string is written as:

Where,

k = spring stiffness constant

Newton’s 3rd law says that the spring willexert a force F in reaction to the force


exert a force FS in reaction to the forceFP such that:

Eq. 6.11 (spring equation) is known as Hooke’s law

FS is the force the spring exerts opposite toits displacement in an attempt to restore its original length

(6-11)

Elastic potential energy is:


(6-12)

When the work done by a force does NOT depend on the path taken, this FORCE is said to be CONSERVATIVE.

Example:

1) Gravitational force: F = mg ; W = mgh (h = vertical displacement)

2) Elastic force: F = -kx ; W = ½ kx 2

Note: If an object starts at a given point and returns to the same point, the net

Conservative and NonConservative and Non--conservative Forcesconservative Forces

Note: If an object starts at a given point and returns to the same point, the net work done is ZERO if the forces acting on it are conservative.

Example:

You lift an object by a vertical displacement h: W1 = mgh

You move the same back to its original position: W2 = mg(-h) = -mgh

Net work done on the object: Wnet = W1 + W2 = mgh + (-mgh) = 0

When the work done by a force DOES depend on the path taken, this FORCE is said to be NON-CONSERVATIVE.

Example:

Friction forces � the work done depends not only on the starting and ending points, but also on the path taken


In general, forces that change as the direction of displacement changes are non-conservatives.

We have seen that the potential energy associated to a force is the energy stored at some point. Gravitational potential energy is an example.

� This only makes sense if this energy can be uniquely defined at a given point (otherwise you would have two potential energies associate to a force at the same point � two different abilities of an object at that particular point to do work �this cannot be!).

However, we just stated that the work done by NON-CONSERVATIVE forces do


However, we just stated that the work done by NON-CONSERVATIVE forces do depend on the path taken

� It does NOT make sense to define potential energy associate to non-conservative forces as you would have more than one potential energy defined at a given point.

� Non-conservative forces do NOT have a potential energy associated to them.

Thus, potential energy can only be defined for cons ervative forces.

General form of the work-energy principle

Observing the two previous slides, we can now generalize the work-energy principle to include the potential energy.

Recalling : We previously wrote � Wnet = ∆KE

Now consider an object subject to both non-conservative and conservative forces. The net work done is:


Where,

General form of the work-energy principle

But, from the work-energy principle, we know that:

We have also seen that the work done by a conservative force, WC , can be written in terms of the potential energy:


(6-13)

terms of the potential energy:

We can then rewrite eq. 6-13 as:

Eq. 6-14 tells us that the work done by non-conserv ative forces is equal to the total change in kinetic and potential energies.

(6-14)

Dissipative energy

The work done by non-conservative force is NOT used to move an object

� The energy used to do this work is dissipated � it will change the total mechanical energy in the system.

Non-conservative forces are also known as DISSIPATIVE FORCES.

Example: When I push my finger on a surface, the initial mechanical energy in my


Example: When I push my finger on a surface, the initial mechanical energy in my fingers is the sum of the potential and kinetic energies .

� But then friction acts against my finger’s motion.

� The work done by the friction force is not used to move your fingers.

� Instead, it tries to slow it down by dissipating part of the total energy available

� This energy is dissipated in the form of thermal energy (heat) � you can feel your fingers getting warmer

If ONLY conservatives forces are present , eq. 6.14 can be written as

or

Mechanical Energy and its ConservationMechanical Energy and its Conservation

We can now define the total mechanical energy stored on a object at certain instant of time as:

And then rewrite eq. 6-15, yielding the equation of conservation of energy:

(6-15)

(6-16)

(6-17)

Eq. 6.17 states the following:

“ If only conservative forces are acting, the TOTAL MECHANICAL ENERGY of a system neither increases nor decreases in any proce ss. It stays constant �� it is CONSERVED”

POWER:

Power is the rate at which work is done, or the rate at which energy is

Mechanical Energy and its ConservationMechanical Energy and its Conservation

Power is the rate at which work is done, or the rate at which energy is transformed .

In the SI system the unit of power is: 1 hp = 746 W

In general, if an object moves with average velocity , the power can be written as

(6-18)

(6-19)

Kinetic Energy, and the Work Energy PrincipleKinetic Energy, and the Work Energy Principle

Problem 6-36 (textbook): In the high jump, Fran’s kinetic energy is transformed into gravitational potential energy without the aid of a pole. With what minimum speed must Fran leave the ground in order to lift her center of mass 2.10 m and cross the bar with a speed of 0.70 m/s ?


Problem 6-36:

We assume that all the forces on the jumper are conservative, so that the mechanical energy of the jumper is conserved.

Subscript 1 represents the jumper at the bottom of the jump, and subscript 2 represents the jumper at the top of the jump.

Call the ground the zero location for PE ( y = 0). We have:

Solve for v1, the speed at the bottom.

1 0y =

2 0.70 m sv =2 2.10 my =

( ) ( )( )

2 2 2 21 1 1 11 1 2 2 1 2 22 2 2 2

22 2

1 2 2

0

2 0.70 m s 2 9.80 m s 2.10 m 6.45 m s

mv mgy mv mgy mv mv mgy

v v gy

+ = + → + = + →

= + = + =


Problem 6-76 (textbook) An airplane pilot fell 370 m after jumping from an aircraft without his parachute opening. He landed in a snowbank, creating a crater 1.1 m deep, but survived with only minor injuries. Assuming the pilot’s mass was 78 kg and his terminal velocity was 35 m/s, estimate

(a) the work done by the snow in bringing him to rest;

(b) the average force exerted on him by the snow to stop him;

(c) the work done on him by air resistance as he fell.(c) the work done on him by air resistance as he fell.


Problem 6-76 :

(a)

Use conservation of energy, including the work done by the non-conservative force of the snow on the pilot. Subscript 1 represents the pilot at the top of the snowbank, and subscript 2 represents the pilot at the bottom of the crater. The bottom of the crater is the zero location for PE ( y = 0 ) . We have

v = 35 m/s , y = 1.1 m, v = 0, y = 0v1 = 35 m/s , y1 = 1.1 m, v2 = 0, y2 = 0

Solve for the non-conservative work:

2 21 1NC 1 2 NC 1 1 2 22 2

W E E W mv mgy mv mgy+ = → + + = + →

( )( ) ( )( )( )22 21 1NC 1 12 2

4 4

78 kg 35 m s 78 kg 9.8 m s 1.1 m

4.862 10 J 4.9 10 J

W mv mgy= − − = − −

= − × ≈ − ×


Problem 6-76 :

(b)

The work done by the snowbank is done by an upward force, while the pilot moves down.

o

N C sn ow sn owco s 18 0 W F d F d= = − →

44 4N C

snow

4.862 10 J4.420 10 N 4.4 10 N

1.1 m

WF

d

− ×= − = − = × ≈ ×


Problem 6-76 :

(c)

To find the work done by air friction, another non-conservative force, use energy conservation including the work done by the non-conservative force of air friction. Subscript 1 represents the pilot at the start of the descent, and subscript 2 represents the pilot at the top of the snowbank. The top of the snowbank is the zero location for PE ( y = 0 ) . We have

v1 = 0 m/s , y1 = 370 m, v2 = 35 m/s, y2 = 0.

Solve for the non-conservative work.

( )( ) ( )( )( )

2 21 1NC 1 2 NC 1 1 2 22 2

22 21 1NC 2 12 2

5 5

78 kg 35 m s 78 kg 9.8 m s 370 m

2.351 10 J 2.4 10 J

W E E W mv mgy mv mgy

W mv mgy

+ = → + + = + →

= − = −

= − × ≈ − ×


Problem 6-63 (textbook): A driver notices that her 1150-kg car slows down from 85 Km/h to 65 Km/h in about 6.0 s on the level when it is in neutral. Approximately what power (watts and hp) is needed to keep the car traveling at a constant 75 km/h?


Problem 6-63

The energy transfer from the engine must replace the lost kinetic energy. From the two speeds, calculate the average rate of loss in kinetic energy while in neutral.

( ) ( ) ( )

1 2

2 22 2 51 1 12 12 2 2

1m s 1m s85 km h 23.61m s 65 km h 18.06 m s

3.6 km h 3.6 km h

1150 kg 18.06 m s 23.61m s 1.330 10 J

v v

KE mv mv

= = = =

∆ = − = − = − ×

Note now that 75 Km/h is the average between the car’s initial and final speeds:

the is needed from the engine.

( ) ( ) ( )

( )2 12 2 2

54 4

1150 kg 18.06 m s 23.61m s 1.330 10 J

1.330 10 J 1 hp2.216 10 W , or 2.216 10 W 29.71 hp

6.0 s 746 W

KE mv mv

WP

t

∆ = − = − = − ×

×= = = × × =

⇒⇒⇒⇒ We can use to conclude

4 12.2 10 W or 3.0 10 hp× ×

chapter 6 – work and energyuregina.ca/~barbi/academic/phys109/2010/notes/lecture-16.pdf ·...

Documents